Chapter 8, Problem 119AP

Calculate $\Delta Hº$ for the reaction

${\text{H}}_{\text{2}}\left(g\right)+{\text{I}}_{\text{2}}\left(g\right)\to 2\text{HI}\left(g\right)$

Using (a) Equation 8.4 and (b) Equation 5.19, given that $\Delta H_{\text{f}}^{º}{\text{ for I}}_{\text{2}}$ (g) is $61.0\text{ kJ/mol}$

Interpretation Introduction

Interpretation:

The standard enthalpy of the reaction is to be calculated.

Concept Introduction:

The standard enthalpy for a reaction is the value of enthalpy at standard conditions that is under room temperature and pressure.

The standard enthalpy of reaction is to be determined using the equation given below:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{°}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{°}\left(\text{reactants}\right).$

Here, the stoichiometric coefficients are represented by m for reactants, and n for products, the enthalpy of formation at standard conditions is represented by $\Delta H_{\text{f}}^{°}$.

The value of enthalpy of formation of an elementis zero at its most stable state.

The enthalpy of reaction can be calculated as:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum BE}\left(\text{reactants}\right)-{\displaystyle \sum BE}\left(\text{products}\right)$.

Here, BE represents the average value of bond enthalpy.

Answer to Problem 119AP

Solution:

(a) $-9.2\text{ kJ}/\text{mol}$

(b) $-9.2\text{ kJ}/\text{mol}$

Explanation of Solution

a) $\Delta H{°}_{\text{rxn}}={\displaystyle \sum BE}\left(\text{reactants}\right)-{\displaystyle \sum BE}\left(\text{products}\right).$

The given reaction is as follows:

${\text{H}}_2\left(g\right)+{\text{I}}_2\left(g\right)\to 2\text{ HI}\left(g\right).$

The enthalpy of reaction can be calculated as:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum BE}\left(\text{reactants}\right)-{\displaystyle \sum BE}\left(\text{products}\right).$

From table 8.6:

$\begin{array}{c}BE\left[\text{H}-\text{H}\right]=436.4\text{ kJ}/\text{mol;}\\ BE\left[\text{H}-\text{I}\right]=298.3\text{ kJ}/\text{mol;}\\ BE\left[\text{I}-\text{I}\right]=151\text{ kJ}/\text{mol}\text{.}\end{array}$

Calculate the standard enthalpy of given reactionas follows:

$\Delta H{°}_{\text{rxn}}=\left\{BE\left[\text{H}-\text{H}\right]+BE\left[\text{I}-\text{I}\right]\right\}-\left\{BE\left[\text{H}-\text{I}\right]\right\}.$

Substitute, $436.4\text{ kJ}/\text{mol}$ for $BE\left[\text{H}-\text{H}\right]$, $298.3\text{ kJ}/\text{mol}$ for $BE\left[\text{H}-\text{I}\right],$ and $151\text{ kJ}/\text{mol}$ for $BE\left[\text{I}-\text{I}\right]$ in the above equation.

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[436.4\text{ kJ}/\text{mol}+151\text{ kJ}/\text{mol}\right]-\left[\text{2}\left(298.3\text{ kJ}/\text{mol}\right)\right]\\ =587.4\text{ kJ}/\text{mol}-596.6\text{ kJ}/\text{mol}\\ =-9.2\text{ kJ}/\text{mol}\text{.}\end{array}$

Hence, the enthalpy of reaction from bond enthalpy value is $-9.2\text{ kJ}/\text{mol}\text{.}$

b) $\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{\text{°}}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{°}\left(\text{reactants}\right)$

The given reaction with its enthalpy of reaction is:

${\text{H}}_2\left(g\right)+{\text{I}}_2\left(g\right)\to 2\text{HI}\left(g\right).$

The standard enthalpy of reaction can be calculated as:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{\text{°}}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{°}\left(\text{reactants}\right).$

From appendix 2,

$\begin{array}{c}\Delta H{°}_{\text{f}}\left[\text{HI}\left(g\right)\right]=25.09\text{ kJ}/\text{mol;}\\ \Delta H{°}_{\text{f}}\left[{\text{I}}_2\left(g\right)\right]=61\text{ kJ}/\text{mol;}\\ \Delta H{°}_{\text{f}}\left[{\text{H}}_2\left(g\right)\right]=0\text{ kJ}/\text{mol}\text{.}\end{array}$

Oxygen is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy of given reaction as follows:

$\Delta H{°}_{\text{rxn}}=\left\{2\Delta H{°}_{\text{f}}\left[\text{HI}\left(g\right)\right]\right\}-\left\{\Delta H{°}_{\text{f}}\left[{\text{H}}_2\left(g\right)\right]+\Delta H{°}_{\text{f}}\left[{\text{I}}_2\left(g\right)\right]\right\}.$

Substitute $25.09\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[\text{HI}\left(g\right)\right]$, $61\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{I}}_2\left(g\right)\right],$ and $0\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{H}}_2\left(g\right)\right]$ in the above equation.

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\left(2\right)\left(25.9\text{ kJ}/\text{mol}\right)\right]-\left[\left(0\text{ kJ}/\text{mol}\right)+61\text{ kJ}/\text{mol}\right]\\ =51.8\text{ kJ}/\text{mol}-61\text{ kJ}/\text{mol}\\ =-9.2\text{ kJ}/\text{mol}\text{.}\end{array}$

Hence, the enthalpy of reaction is $-9.2\text{ kJ}/\text{mol}$.